{"id":3634,"date":"2018-02-13T12:57:10","date_gmt":"2018-02-13T12:57:10","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3634"},"modified":"2024-08-20T23:13:23","modified_gmt":"2024-08-20T23:13:23","slug":"resolucao-comentada-forcas-no-movimento-circular","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/forcas-no-movimento-circular\/resolucao-comentada-forcas-no-movimento-circular\/","title":{"rendered":"FOR\u00c7AS NO MOVIMENTO CIRCULAR – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada <\/b><\/span><\/span><\/span><\/p>\n

FOR\u00c7AS NO MOVIMENTO CIRCULAR<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe na figura da direita que o pequeno bloco preso \u00e0 extremidade do fio, chamado de \u201cpeso\u201d descreve uma circunfer\u00eancia em um plano horizontal\u00a0 —\u00a0 nesse caso, surge sobre ele uma for\u00e7a resultante centr\u00edpeta\u00a0 —\u00a0 a for\u00e7a que faz a garrafa subir \u00e9 a tra\u00e7\u00e3o que age sobre a garrafa produzindo seu deslocamento vertical para cima e, ela pode ser considerada como uma parcela da for\u00e7a resultante centr\u00edpeta\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Tanto no ponto mais alto (A) como no ponto mais baixo (B) as for\u00e7as que agem sobre o carv\u00e3o \u00a0s\u00e3o, seu peso\u00a0\"\"\u00a0constante, vertical e para baixo e a for\u00e7a de compress\u00e3o\u00a0\"\"\u00a0que o carv\u00e3o troca com a superf\u00edcie interna da lata para baixo em (A) e para cima em (B).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Em (A), ponto mais alto, a for\u00e7a resultante centr\u00edpeta\u00a0\u00a0\"\", que \u00e9 dirigida sempre para o centro da circunfer\u00eancia, no caso, para baixo, \u00e9 a soma vetorial de\u00a0\"\"\u00a0com\u00a0\"\", de intensidade\u00a0 — \u00a0F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=P + N\u00a0 —\u00a0 m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=m.g + N\u00a0 — \u00a0N=m(g + V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R).\u00a0\u00c9 o mesmo resultado obtido quando um ve\u00edculo passa por uma lombada ou quando um piloto de avi\u00e3o se encontra na parte mais alta de um looping vertical.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Substituindo os dados do exerc\u00edcio na express\u00e3o V=\u221a(Rg)=\u221a(0,8.10)\u00a0 —\u00a0 V=\u221a8\u00a0 —\u00a0 V=2\u221a2m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na parte mais baixa da trajet\u00f3ria as for\u00e7as que agem sobre a lata s\u00e3o seu peso\u00a0\"\"\u00a0(vertical e para baixo) e a for\u00e7a de tra\u00e7\u00e3o na<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

corda\u00a0\"\"\u00a0(vertical e para cima)\u00a0 — a intensidade de\u00a0\"\"\u00a0deve ser maior que a intensidade de\u00a0\"\", pois a for\u00e7a resultante centr\u00edpeta\u00a0\"\" deve ser vertical e para cima, ou seja, dirigida para o centro da circunfer\u00eancia\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 n=4 rota\u00e7\u00f5es\u00a0 —\u00a0 \u2206t=1s\u00a0 —\u00a0 W=n.360\/\u2206t\u00a0 —\u00a0 W=4.360\/2\u00a0 —\u00a0 W=720\u00a0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0\u00a0 —\u00a0 \u00a0m = 1.000 kg\u00a0 —\u00a0 \u00a0H = 30 m\u00a0 —\u00a0 \u00a0d = 20 m\u00a0 —\u00a0 r = 10 m\u00a0 —\u00a0 observe a figura <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

abaixo\u00a0 —\u00a0 c\u00e1lculo da velocidade no ponto B pela conserva\u00e7\u00e3o de energia mec\u00e2nica\u00a0 —\u00a0\u00a0E<\/b><\/span><\/span><\/span>MA<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>MB<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—mgH=mgd + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2h(H \u2013 d)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=20.(30 \u2013 20)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=200\u00a0 —\u00a0 no ponto B, a resultante das for\u00e7as que agem sobre o carro s\u00e3o radiais, portanto a acelera\u00e7\u00e3o \u00e9 centr\u00edpeta\u00a0 —\u00a0 a=ac=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R=200\/10\u00a0 —\u00a0a=20m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

No ponto B, a resultante \u00e9 centr\u00edpeta e a for\u00e7a vertical que o trilho exerce no carro \u00e9 a normal\u00a0\"\"\u00a0\u00a0—\u00a0 no ponto B a for\u00e7a resultante centr\u00edpeta tem dire\u00e7\u00e3o vertical e sentido para baixo (centro da circunfer\u00eancia)\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=N + P=ma\u00a0 —\u00a0 N + P=ma\u00a0 —\u00a0 N=ma \u2013 mg\u00a0 —\u00a0 N=1.000(20 \u2013 10)\u00a0 —\u00a0\u00a0N=10.000N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Conforme o diagrama anexo, as for\u00e7as que agem no carro s\u00e3o o peso\u00a0\"\"\u00a0e a normal\u00a0\"\". Como o movimento \u00e9 circular e uniforme, a resultante dessas for\u00e7as \u00e9 centr\u00edpeta (radial),\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

tg\u03b1 = F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/P=m.a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/mg\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=g.tg\u03b1\u00a0 —\u00a0 como a e g s\u00e3o constantes, a acelera\u00e7\u00e3o centr\u00edpeta (radial, dirigida para o centro) tem m\u00f3dulo constante\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O objeto solto bem como o pr\u00f3prio sat\u00e9lite est\u00e1 sujeito \u00e0 for\u00e7a gravitacional terrestre e logo ambos t\u00eam peso\u00a0 —\u00a0 admitindo que o movimento do sat\u00e9lite e do corpo s\u00e3o circulares e uniformes, a acelera\u00e7\u00e3o centr\u00edpeta ser\u00e1 a acelera\u00e7\u00e3o gravitacional\u00a0 —\u00a0 a alternativa c \u00e9 falsa, pois se uma pessoa estiver solta dentro da nave ela n\u00e3o experimenta sensa\u00e7\u00e3o de peso, mas est\u00e1 sujeito \u00e0 acelera\u00e7\u00e3o gravitacional\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O dinam\u00f4metro indica a intensidade da for\u00e7a de tra\u00e7\u00e3o\u00a0\"\"\u00a0no fio em cada caso\u00a0 —\u00a0 na figura (I), <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

supondo o corpo em equil\u00edbrio, a intensidade da for\u00e7a de tra\u00e7\u00e3o\u00a0\"\"\u00a0ser\u00e1 a mesma que a do peso\u00a0\"\"\u00a0\u00a0—\u00a0na figura (II), p\u00eandulo c\u00f4nico com o corpo girando com velocidade angular constante\u00a0 —\u00a0 colocando as for\u00e7as do lado direito, a soma vetorial de\u00a0\"\"\u00a0com\u00a0\"\"\u00a0fornecer\u00e1 a for\u00e7a resultante centr\u00edpeta\u00a0\"\"\u00a0\u00a0—\u00a0\u00a0\"\"\u00a0=\u00a0\"\"\u00a0+\u00a0\"\"\u00a0\u00a0—\u00a0 observe na figura (II) no tri\u00e2ngulo hachurado da direita que T=P\/cos\u03b8 —\u00a0 como cos\u03b8<1\u00a0 —\u00a0 T>P\u00a0 —\u00a0 a intensidade de\u00a0\"\", que representa a indica\u00e7\u00e3o do dinam\u00f4metro \u00e9 maior que o peso do corpo\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para que o astronauta permane\u00e7a na mesma posi\u00e7\u00e3o \u201ccolado\u201d \u00e0 nave, ele deve ter a mesma velocidade angular W que ela e tamb\u00e9m o mesmo per\u00edodo T (tempo que ela demora para efetuar uma volta completa), que \u00e9 de 20s.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

T=20s\u00a0 —\u00a0 W=2\u03c0 \/T\u00a0 —\u00a0 W=2\u03c0 \/20= \u03c0 \/10\u00a0 —\u00a0 W= \u03c0 \/10rad\/s\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=m.W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.R\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=70.( \u03c0 \/10)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.10\u00a0 —\u00a0\u00a0F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=7\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A for\u00e7a resultante centr\u00edpeta\u00a0\u00a0\"\", que \u00e9 dirigida sempre para o centro da circunfer\u00eancia, no caso, <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

para cima, \u00e9 a diferen\u00e7a vetorial entre\u00a0\"\"\u00a0e\u00a0\u00a0\"\", de intensidade\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=N \u2013 P\u00a0 —\u00a0\u00a0 m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R= N \u2013 m.g\u00a0 —\u00a0\u00a0N=m.g + mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 N=50.10 + 50.100\/2\u00a0 —N=3.000N\u00a0 —\u00a0 N= 3,0.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo o movimento circular a for\u00e7a resultante sobre a mosca \u00e9 a centr\u00edpeta de dire\u00e7\u00e3o radial, sentido para o centro da circunfer\u00eancia e intensidade F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 como m e V s\u00e3o constantes, F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0e R s\u00e3o inversamente proporcionais, ou seja, quanto menor \u00e9 o raio R, maior ser\u00e1 a intensidade da for\u00e7a resultante centr\u00edpeta\u00a0\"\"\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o Comentada FOR\u00c7AS NO MOVIMENTO CIRCULAR 01-\u00a0Observe na figura da direita que o pequeno bloco preso \u00e0 extremidade do fio, chamado de \u201cpeso\u201d descreve uma circunfer\u00eancia em um plano horizontal\u00a0 —\u00a0 nesse caso, surge sobre ele uma for\u00e7a resultante centr\u00edpeta\u00a0 —\u00a0 a for\u00e7a que faz a garrafa subir \u00e9 a tra\u00e7\u00e3o que age sobre a garrafa produzindo seu deslocamento vertical<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3632,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-3634","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3634","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=3634"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3634\/revisions"}],"predecessor-version":[{"id":10668,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3634\/revisions\/10668"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3632"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3634"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}